# Quant Boosters - Vikas Saini - Set 1

• let 6th month he had a sale of z.
6435+6927+6855+7230+6562+z = 6 x 6500
= > 34009 + z = 39000
= > z = 4991.

Short approach :-
let's take 6500 as base.
6500-(-65+427+355+730+62)
=6500 - (1509)
= 4991.

• Q6) The average weight of boys in a school is 47 and girls is 42. If there are 6 more boys than girls and average weight of all class is 44.8. How many students are in the class?

• let no of girls = a
no of boys = a+6
47(a+6)+42a = 44.8(a+a+6)
47a+42a+282 = 44.8(2a+6)
89a + 282 = 89.6a + 268.8
0.6 a = 13.2
a = 22.
a+6 = 28.
total = 22+28 = 50.

Short approach :-
average of all class=44.8.
ratio of boys & girls= (47-44.8 ):(44.8-42)=2.2:2.8.
(2.8-2.2)x=6
x=10.
total students= (2.2+2.8 )x10=50

• Q7) Average weight of 11 players is increased by 1kg. When one player of team weighing 55kg is replaced by new player. Find weight of new player

• Average weight of each player = a.
weight of new player = k.
11a - 55 + k = 11(a+1)
k - 55 = 11
k = 66.

Short approach :-
weight of new player = 55 + 1 x 11 = 66.

• Q8) The average of 71 results is 48. If the average of first 59 results is 46 and that of the last 11 is 52 then find the 60th result.

• 71 x 48 = 59 x 46 + 11 x 52 + 60th
= > 60th value = 71 x 48 - 59 x 46 - 11 x 52.
= > 60th value = 122.

Short approach :-
We take 48 as base.
59 results average is -2 (from base)
11 results average is +4 (from base)
So 59 x (-2) + 11 x 4 = -74.
To get 60th value this value should be decrease from base.
Hence 60th value = 48 - (-74) = 122.

• Q9) The average of 50 numbers is 38. If two numbers, namely 45 & 55 are discarded then what is the average of remaining numbers ?

• Total = 50 x 38 = 1900.
new total = 1900-45-55 = 1800.
new average = 1800 / 48 = 37.5

Short approach :-
The average of 50 numbers is 38.
so if we discarded two number then to maintain average of 38,value should be decreased by 2 x 38 = 76.
but here value decreased by 100.
So 100-76 = 24.
There is 48 numbers.
so every number value will be decreased by 24/48 = 0.5
new average = 38 - 0.5 = 37.5.

• Q10) The average runs of Rohit & Dhawan is 280. The average runs of Virat & Rahane is 200. What is the average of each of them.

• Rohit + Dhawan = 2 x 280 = 560.
Virat + Rahane = 2 x 200 = 400.
Total runs = 560 + 400 = 960.
Average = 960 / 4 = 240.

Short approach :-
Take average of any two
let we take Rohit & Dhawan,which is nothing but 280.(base)
now average of virat & rahane 200.
which is -80 (from base)
Average = 280 - (80/2)
= 280 - 40
= 240.

We can solve it by take 200 too..
base = 200.
average of Rohit & Dhawan is +80(from base).
average = 200 + (80/2)
= 240.
Here divided by 2, because that is average of two person..

• Q11) The average weight of a class of 29 students is 40 kg. If the weight of teacher be included,the average rises by 500gm. What is the weight of the teacher ?

• previous weight of class = 29 x 40 = 1160 kg.
new weight (including teacher) = 30 x 40.5 = 1215.
Teacher's weight = 1215 - 1160 = 55 kg.

Short approach :-
base = 40.
including teacher average increased by 0.5 kg.
Teacher's weight = 40 + 30 x 0.5
= 40 + 15
= 55 kg.

• Q12) By selling 18 oranges profit is equal to SP of 3 oranges. Find profit percentage.

• Sp of 1 orange = x.
cp of 1 orange = y.
18x – 18y = 3x
= > 18 y = 15 x.
Profit percentage = (3x/18y)x100
= (3x/15x)x100
= 20%.

Short approach :-
Profit percentage = [profit (quantity) / (Total quantity – profit)] x 100
= [(3)/(18-3)] x 100
= [3/15] x 100
= 20%

• Q13) By selling 18 chocolates, vendor loss selling price of 2 chocolates. Find loss percentage.

• sp of 1 chocolate = x
cp of 1 chocolate = y
18 y – 18 x = 2x.
18 y = 20 x.
Loss percentage = (2x/18 y) x 100
= (2x/20x)x100
= 10%.

Short approach :-
Loss percentage = [Loss (quantity) / (Total quantity + loss) ] x 100
= [ 2 / (18+2) ] x 100
= 10%.

• Q14) On selling an article for rs 576 a trader losses 4%. In order to gain 25/6 % , he must sell article for
a. 655 rs
b. 676 rs
c. 625 rs
d. 600 rs.

• Cost price = [576 / (100 – 4)] x 100 = 600 rs.
Selling price = 600 x (1 + 25/(6 x 100) = 625.

Short approach :-
Direct formula to find sp in such problems
= previous selling price x 100 x (1+ gain percent)/(1 – loss percent)
= 576 x 100 x (1+25/600)/(1 – 4/100)
= 625 rs.

• Q15) A goldsmith has 361 rings of gold. He sells some of them at a loss of 4% and rest at a profit of 15%. Overall profit is 8%. Find no of rings sold at profit of 15%.

61

62

46

63

31

60

61

61